Problem 25 - Calculus Early Transcendentals

Question

Find the mass of the following objects with the given density functions. The solid cone \(D=\\{(r, \theta, z): 0 \leq z \leq 6-r, 0 \leq r \leq 6\\}\) with density \(\rho(r, \theta, z)=7-z\)

Step-by-Step Solution

Answer

Answer: The mass of the solid cone is \(930\pi\) units.

Step 1: Set up the triple integral with the given density function

We will first set up the integral to find the mass of the cone: $$ Mass = \iiint_\mathcal{D} \rho(r, \theta, z) \, dv $$ Here, \(dv\) represents the volume differential in polar coordinates, which is given as $$ dv = r\,dz\,dr\,d\theta. $$

Step 2: Determine the limits of integration

To find the limits of integration, let's analyze the cone's constraints: - \(0 \leq z \leq 6-r\): This tells us that z goes from 0 to \(6-r\). - \(0 \leq r \leq 6\): This tells us that r goes from 0 to 6. - Since the cone is closed and makes a full circle, the limits for \(\theta\) will be \(0\) to \(2\pi\). Now, we can combine these limits with our integral: $$ Mass = \int_{0}^{2\pi} \int_{0}^{6} \int_{0}^{6-r} \rho(r, \theta, z) \, r\,dz\,dr\,d\theta $$

Step 3: Substitute the density function into the integral

We are given the density function \(\rho(r, \theta, z) = 7-z\). Substituting this into the integral, we get: $$ Mass = \int_{0}^{2\pi} \int_{0}^{6} \int_{0}^{6-r} (7-z) \, r\,dz\,dr\,d\theta $$

Step 4: Integrate with respect to z

First, we will integrate with respect to z: $$ Mass = \int_{0}^{2\pi} \int_{0}^{6} \left[\int_{0}^{6-r} (7z - z^2)r\,dz\right] \, dr\,d\theta $$ Integrating, we get: $$ Mass = \int_{0}^{2\pi} \int_{0}^{6} (42r - 12r^2 + r^3) \, dr\,d\theta $$

Step 5: Integrate with respect to r

Now, we will integrate with respect to r: $$ Mass = \int_{0}^{2\pi} \left[\int_{0}^{6} (42r - 12r^2 + r^3) \, dr\right] \, d\theta $$ Integrating, we get: $$ Mass = \int_{0}^{2\pi} (756 - 372 + 81) \, d\theta $$ Simplifying, we get $$ Mass = \int_{0}^{2\pi} 465\, d\theta $$

Step 6: Integrate with respect to θ

Finally, we will integrate with respect to θ: $$ Mass = \int_{0}^{2\pi} 465\, d\theta $$ Integrating and simplifying, we get: $$ Mass = 930\pi $$ Thus, the mass of the solid cone is \(930\pi\) units.